Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes.
Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?
Answer
(D)
Explanation
IP fragments the whole UDP segment, header included: \(8 + 7488 = 7496\) bytes of IP data.
A fragment has a 20-byte IPv4 header and must fit in 1500 bytes, so it carries at most 1480 data bytes. Since 1480 is a multiple of 8, as fragment offsets require, every fragment except the last carries exactly 1480 bytes.
- Five full fragments carry \(5 \times 1480 = 7400\) bytes.
- The remaining \(7496 - 7400 = 96\) bytes form a sixth fragment of \(96 + 20 = 116\) bytes.
Leaving out the UDP header gives 88 data bytes (108 bytes with the IP header), the trap in options B and C.
Answer: D